Concatenating n with n + 1

Question

Introduction

OEIS sequence A127421 is the sequence of numbers whose decimal expansion is a concatenation of 2 consecutive increasing non-negative numbers. Put simply, every number in the sequence is formed by putting together n with n+1 for some non-negative, integer value of n. The first several terms are:

1, 12, 23, 34, 45, 56, 67, 78, 89, 910, 1011, 1112, 1213, 1314, 1415, 1516, 1617, 1718, 1819, 1920, 2021, 2122, 2223, 2324, 2425, 2526, 2627, 2728, 2829, 2930, 3031, 3132, 3233, 3334, 3435, 3536, 3637, 3738, 3839, 3940, 4041, 4142, 4243, 4344, 4445, 4546, …

Challenge

Given a single positive integer n, print the first n entries of OEIS sequence A127421 in increasing order.

• Input and output can be in any acceptable format. Strings or numbers are fine for output.
• Leading zeroes are not permitted.
• Either a full program or function is permitted.
• For the purposes of this challenge, n will be positive and under 100.
• Standard loopholes are disallowed by default.
• This question is code golf, so lowest byte-count wins.

• Here is some sample input and output:

  1 => 1
  2 => 1, 12
  3 => 1, 12, 23
  10 => 1, 12, 23, 34, 45, 56, 67, 78, 89, 910
  

If you have any questions, don't hesitate to ask. Good luck.

P.S this is my first challenge, so hopefully this all makes sense.

EDIT: Removed output restriction to allow numbers or strings.

Answer 1

Jelly, 3 bytes

ŻVƝ

A monadic link accepting an integer which yields a list of integers

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How?

ŻVƝ - Link: integer       e.g. 59
Ż   - zero-range               [0,1,2,3,4,5,6, ... ,58,59]
  Ɲ - apply to each pair: i.e: [0,1] or [5,6]  or  [58,59]
 V  -   evaluate* jelly code   1     or 56     or  5859
    -                       -> [1,12,23,45,56, ... 5859]

* When given a list V actually joins the Python string values and evaluates that
  ...so e.g.: [58,59] -> ['58','59'] -> '5859' -> 5859

Answer 2

R, 32 bytes

strtoi(paste0((x=1:scan())-1,x))

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Outgolfed by MickyT, so go upvote that answer!

Answer 3

Python 3, 39 bytes

f=lambda n:1//n or f'{f(n-1)} {n-1}{n}'

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Answer 4

Haskell, 38 37 bytes

f n=("":)>>=zipWith(++)$show<$>[1..n]

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Thanks to Cat Wizard for a byte!

Answer 5

Perl 6, 19 18 bytes

{(^$_ Z~1..$_)X+0}

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Anonymous code block that zips the range 0 to n-1 with 1 to n using the concatenation operator, then adds 0 to every element to force it to a number and remove leading 0s.

Answer 6

Cubix, 19 bytes

I.1.W)>OSo;u.uO;@!-

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This wraps onto the cube as follows

    I .
    1 .
W ) > O S o ; u
. u O ; @ ! - .
    . .
    . .

Watch It Run

Got a little room to play with yet, but at the moment

• W redirect to the top face heading down
• I1> set up the stack with the input and 1 then redirect into the main loop
• OSo;u output the top of stack, add space to stack, output, remove and uturn
• -!@;Ou) subtract TOS from input, if 0 halt else pop result, output TOS, uturn and increment TOS. Back into the main loop.

Answer 7

R, 30 29 bytes

An extra byte thanks to @Giuseppe

10^nchar(n<-1:scan())*(n-1)+n

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A mostly mathematical solution, except for using nchar() rather than floor(log10()). I was really surprised that it came in shorter than the string version.

Answer 8

APL (Dyalog), 13 12 bytes

1 byte saved thanks to @FrownyFrog

(⍎⍕,∘⍕1∘+)¨⍳

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Answer 9

Python 2, 42 41 bytes

f=lambda n:n-1and f(n-1)+[`n-1`+`n`]or[1]

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Recursive function that returns a mixed list of strings and integers

Answer 10

Brachylog, 6 bytes

⟦s₂ᶠcᵐ

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Explanation

⟦         Range: [0, …, Input]
 s₂ᶠ      Find all substrings of length 2
    cᵐ    Map concatenate

Answer 11

Haskell, 34 bytes

f n="1":[show=<<[i-1,i]|i<-[2..n]]

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Answer 12

Blossom, 88 bytes

rule e<int x>[1(x)]=>[1(x-1),2(str(x)+str(x+1))];rule c[1(0)]=>[];e!c

Blossom is a graph programming language I'm working on. It can only take graphs as inputs, so this programme expects a graph comprising a single node with its label an integer. It returns a graph of connected edges to form the closest to an array I can get, and the resultant graph is printed to output.

An unminified version of the code is this:

rule expand <int x>
    [ 1 (x) ]
 => [ 1 (x-1), 2(str(x)+str(x+1)) ]
where x > 0;

rule clean
    [ 1 (0) ]
 => [];

expand! clean

It defines two rules: one called expand, which (while there is a node with an integer-valued label in the current graph) creates another node with its increment concatenated, and lowers the value. This rule also has the condition that x is greater than 0.

The ! executes this rule for as long as it can be applied on the graph, so in this case it will execute until x is 0. And then the clean rule removes this 0 node.

Blossom was not made for golfing, but it doesn't do too badly, I don't think., considering what it is. There currently isn't really an easy way for people to test blossom code (and the interpreter I'm working on at the moment is not quite finished and a little buggy), but this isn't exactly a competing entry!

Answer 13

JavaScript (Node.js), 25 bytes

f=n=>--n?f(n)+','+n+-~n:1

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Answer 14

C (gcc), 44 43 bytes

f(i){i--&&printf(" %2$d%d"+5*!f(i),i+1,i);}

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Answer 15

Shakespeare, 703 bytes

Q.Ajax,.Ford,.Act I:.Scene I:.[enter Ajax and Ford]Ford:Open mind!Scene V:.Ajax:You is the sum of thyself the sum of myself the sum of a big bad fat old red pig a big bad fat old lie!Ford:Open mind!Is you nicer zero?Ajax:If so, you is twice the sum of the sum of twice thyself twice thyself thyself!If so,Let us Scene V!Ford:You a cat!Open heart!Scene X:.Ajax:You is the sum of thyself a pig!Is you worse than a cat?If so,let us Scene C.Remember thyself.You is the sum of the sum of a big old red cute rich cat a big old red cute joy a big old pig!Speak mind!You is a big old red cute rich cat!Speak mind!Recall!Ford:Open heart!You is the sum of thyself a joy!Open heart!Let us Scene X.Scene C:.[exeunt]

try it here

ungolfed version

127421th Night.
Ajax, likes to read the stars.
Ford, someone Ajax can always count on.
Act I:.
Scene I: Ajax reads a star.
[enter Ajax and Ford]
Ford: Open your mind! 
Scene V: Ford counts what ajax has learned.
Ajax: you are the sum of thyself and the sum of myself and the sum of a big bad fat old red pig and a big bad fat old lie!
Ford: Open Your mind! Are you nicer than zero?
Ajax: If so, you are twice the sum of the sum of twice thyself and twice thyself and thyself! 
If so, Let us Scene V!
Ford: You are a cat! Open your heart!

Scene X: Ajax and Ford recall the nights.
Ajax: You are the sum of thyself and a pig! Are you worse than a cat? If so, Let us Scene C.
Remember thyself. 
You are the sum of the sum of a big old red cute rich cat and a big old red cute joy and a big old pig! 
Speak you mind!
You are a big old red cute rich cat! Speak your mind! Recall your finest hour!
Ford: Open your heart! You are the sum of thyself and a joy! Open your heart! Let us Scene X.
Scene C: Fin.
[exeunt]

Answer 16

Groovy, 35 bytes

{(0..<it)*.with{""+it+++it as int}}

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I came up last minute with the idea of using *.with instead of .collect. I have no idea what it+++it parses to but whether it's it++ + it or it + ++it they both do the same thing. I tried to think of a way of getting rid of the < in ..< by turning it into 1..it and decrementing but I don't think it would get any shorter.

Answer 17

Pyth, 9 8 6 bytes

ms+`dh

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Explanation:

       - implicit output
m      - map function with argument d:
  +    -  concatenate
    d  -  argument d
   `   -  to string
     h -  into implicit d + 1
       - into Q (implicit input)

Answer 18

Python 3, 55 48 47 43 bytes

f=lambda n:n-1and f(n-1)+[f"{n-1}{n}"]or[1]

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Recursive function that takes an integer and returns a mixed list of strings and numbers.

Answer 19

Jelly, 4 bytes

ḶżRV

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How it works

ḶżRV  Main link. Argument: n

Ḷ     Unlength; yield [0, ..., n-1].
  R   Range; yield [1, ... n].
 ż    Zipwith; yield [[0, 1], ..., [n-1, n]].
   V  Eval; cast each array to string and evaluate, yielding integers.

Answer 20

Python 2, 44 bytes

for i in range(input()):print`i`*(i>0)+`i+1`

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Answer 21

Japt -m, 6 5 bytes

ó2 ¬n

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As always, know the flags.

Unpacked & How it works

-m       Convert to range and map...

Uó2 q n
Uó2      Construct [U, U+1]
    q    Join
      n  Convert to number

         Implicit output (Array is printed as comma-delimited values)

Answer 22

05AB1E, 6 bytes

>GNJ,N

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Explanation

>G       # for N in [1 ... input]
  N      # push N
   J     # join stack
    ,    # print
     N   # push N (for next iteration)

LεD<ìï would work for same byte count but with list output

Answer 23

APL (Dyalog Classic), 9 bytes

1,2,/⍕¨∘⍳

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Answer 24

Powershell, 27 26 bytes

1.."$args"|%{"$p$_";$p=$_}

-1 byte: thanks AdmBorkBork

Test script:

$f = {
1.."$args"|%{"$p$_";$p=$_}
}

&$f 1
""
&$f 2
""
&$f 3
""
&$f 10
""
&$f 46

Answer 25

Husk, 11 9 bytes

^{-2 thanks to @BMO!}

mSöd+d←dḣ

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Explanation

m          map(                                              )
 S                                               <*>
  ö            (           .).    .         .
   d            fromDecimal
    +                         (++)
     d                             toDecimal
      ←                                      (+1)
       d                                            toDecimal
                                                              .
        ḣ                                                      (\n->[1..n])

mSöd+d←dḣ  map((fromDecimal.).(++).toDecimal.(+1)<*>toDecimal).(\n->[1..n])

Answer 26

C# (Visual C# Interactive Compiler), 103 71 64 56 bytes

---

Golfed Try it online!

i=>{for(int x=0;x<i;)Write($"{(x>0?$",{x}":"")}{++x}");}

---

Ungolfed

i => {
    for( int x = 0; x < i; )
        Write( $"{( x > 0 ? $",{x}" : "")}{ ++x }" );
}

---

Full code

Action<Int32> a = i => {
    for( int x = 0; x < i; )
        Write( $"{( x > 0 ? $",{x}" : "")}{ ++x }" );
    };

Int32[]
    testCases = new Int32[] {
        1,
        2,
        3,
        10,
    };

foreach( Int32[] testCase in testCases ) {
    WriteLine( $" Input: {testCase}\nOutput:" );
    a(testCase);
    WriteLine("\n");
}

---

Older versions:

• v1.2, 64 bytes

  i=>{for(int x=0;x<i;)Write($"{(x>0?$",{x}":"")}{++x}");}
  

• v1.1, 71 bytes

  i=>{for(int x=0;x<i;)System.Console.Write($"{(x>0?$",{x}":"")}{++x}");}
  

• v1.0, 103 bytes

  i=>{for(int x=0;x<i;)System.Console.Write($"{(x>0?",":"")}{x++*System.Math.Pow(10,$"{x}".Length)+x}");}
  

---

Releases

• v1.3 - - 8 bytes - Removed Console thanks again to raznagul
• v1.2 - - 7 bytes - Removed System thanks to raznagul
• v1.1 - -32 bytes
• v1.0 - 103 bytes - Initial solution.

---

Notes

• None

Answer 27

Haskell, 37 bytes

f x=[[y-1|y>1]++[y]>>=show|y<-[1..x]]

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Answer 28

J, 14 bytes

(,&.":>:)"0@i.

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Answer 29

ABAP, 101 bytes

Not really a golfing language, but I'm having a lot of fun with it

WHILE x<w.
CLEAR z.
IF x=1.
WRITE x.
ELSE.
CONCATENATE y x INTO z.
WRITE z.
ENDIF.
y=x.
x=x+1.
ENDDO.

W is the input term, X is the counter from 1, Y is X-1 from the second pass onward, Z is concatenated string.

Answer 30

PHP, 33 32 bytes

while($argv[1]--)echo" $i".++$i;

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Old version

for(;$i<$argv[1];)echo" $i".++$i;     // 33 bytes

Answer 31

Python 2, 41 bytes

lambda l:[`n`[:n]+`n+1`for n in range(l)]

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Answer 32

Javascript, 43 44,46,49,53 bytes

n=>[...Array(n)].map((_,a)=>~~(a+(a+1+"")))

Previous versions :

n=>[...Array(n)].map((_,a)=>~~(a+(""+ ++a)))
n=>[...Array(n)].map((_,a)=>0- -(a+(""+ ++a)))
n=>[...Array(n).keys()].map(a=>0- -(a+(""+ ++a)))
n=>[...Array(n).keys()].map(a=>parseInt(a+(""+ ++a)))

Saved 3 bytes thanks to @Shaggy's solution (which is better than mine) to remove .keys()

Answer 33

Python 3, 44 bytes

lambda n:[f"{j or''}{j+1}"for j in range(n)]

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Answer 34

K (ngn/k), 13 bytes

{.,/$x+!2}'!:

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! generate the list 0 1 ... n-1

{ } is a function with argument x

' applied to each

!2 is 0 1

x+!2 is x, x+1

$ format as strings

,/ concatenate

. evaluate (convert to number)

Answer 35

Canvas, 6 bytes

ŗ²；＋┤］

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Explanation:

{     ]  map over 1..input
 ŗ         convert to string (required as `＋` needs 1 arg to be a string to not add)
  ²;       place a 0-indexed version of the index below TOS
    +      join the two
     ┤     cast to number

ŗ²×┤］ (or even ²×┤］) would work if I didn't try to push the absolute most out of characters and didn't make × - reverse add - do not that when not receiving 2 strings.

Answer 36

cQuents, 11 bytes

=1::($-1)~$

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Way too long for a language that this is supposed to be easy in, guess I should have implemented unary subtract-one and add-one operators.

Explanation

=1            First term in sequence is 1
  ::          Mode: for input n, output first n terms in sequence
              For each term in the sequence:
    ($-1)      Subtract 1 from the current index
         ~$    Concatenate that to the current index

Answer 37

APL (Dyalog Classic), 13 bytes

2{⍎∊⍕¨⍺⍵}/0,⍳

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Answer 38

Perl 5, 24 21 bytes

@DomHastings came up with a way to save 3 bytes

say$i++.$i|0for 1..<>

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Answer 39

Momema, 36 bytes

z0w+1=*0-8*0w00+1*0-8*0-9 10z=+_M-*0

Try it online! Requires the -i interpreter flag.

Explanation

                                                     #  i = 0
d   0            #  d0:  nop                         #  do {
i   (+ 1 =*0)    #  i0:  jump to i(![0])             #    if i {
-8  *0           #       print [0]                   #      print i
i   0            #  i1:  nop                         #    }
0   (+ 1 *0)     #       [0] = [0] + 1               #    i += 1
-8  *0           #       print [0]                   #    print i
-9  10           #       print chr(10)               #    print \n
d   =(+ _M -*0)  #  d1:  jump to d(!!(input - [0]))  #  } while (input - i != 0)

The Momema reference interpreter provides an "interactive mode", enabled by -i, which is intended to be used on the command line.

One of the features it allows is the ability to add holes, denoted by _, which effectively allow one to substitute an integer read from STDIN into an expression. This is already shorter than *-8 (read from memory-mapped I/O location -8), which does the same thing without interactive mode and without displaying a prompt to STDERR.

Crucially, holes can also be named with a sequence of capital letters after the _. Evaluation of named holes is memoized. Input will be read the first time a named hole is evaluated but subsequent evaluations of a hole with the same name will reuse the input number. This means that we can use _M to stand in for "the input", but input will only be read on the first iteration of the loop.

Answer 40

Red, 56 bytes

func[n][prin"1 "repeat i n - 1[prin rejoin[i i + 1" "]]]

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Answer 41

Rust, 56 bytes

|n|(0..n).map(|n|format!("{}{}",n,n+1).parse().unwrap())

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Answer 42

Java, 48 bytes

String f(int n){return--n>0?f(n)+","+n+-~n:"1";}

Port of tsh's JavaScript answer. Try it online here.

Ungolfed:

String f(int n) { // recursive function taking an integer as argument and returning a String
    return --n > 0 ? f(n) // decrement n and recurse if n is still positive after
                   + "," + n // return the result of the recursive call concatenated with n and ...
                   + (- ~n) // ... n+1; writing it as -~n gives it precedence over the concatenation
                   : "1"; // if n is now 0 on the other hand, return "1"
}

Answer 43

MATL, 9 bytes

:"@qV@VhU

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              % implicit input n
:             % range, push 1..n
"             % for loop:
 @            % push for loop index
 q            % decrement
 V            % convert to string (num2str)
 @            % push for loop index
 V            % convert to string (num2str)
 h            % horizontally concatenate
 U            % convert to number (str2num)
              % implicit end of for loop
              % implicit end of program, display stack contents

Answer 44

Panacea, 6 bytes

re
D>j

Explanation:

r        Range [0.. input]
 e       Map each element with the following line
D        Duplicate
 >       Increment
  j      Join digits; multiply by ten and add
         Since this works with integers, it removes the preceding 0.

Answer 45

Japt, 5 bytes

õÈsiY

Try it

---

Explanation

          :Implicit input of integer U
õ         :Range [1,U]
 È        :Map each integer at index Y
  s       :  Convert to string
   iY     :  Prepend index
          :  Implicitly convert back to integer
          :Implicit output of resulting array

---

Alternative

Ç°s+Z

Try it

          :Implicit input of integer U
Ç         :Map each integer Z in the range [0,U)
 °        :  Postfix increment
  s       :  Convert to string
   +Z     :  Append (the now incremented) Z
          :  Implicitly convert back to integer
          :Implicit output of resulting array

Answer 46

J, 20 bytes

".(,&":)/@(,>:)"0@i.

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Answer 47

Forth (gforth), 33 bytes

: f dup 1 ?do i . i 1 .r loop . ;

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Explanation

Loops through all numbers in range and outputs the number twice, the first with a space appended, and the second without.

This results in the first output being appended to the previous output. To fix the last case, the input is outputted so that it can be appended to the last number in the loop.

Code Explanation

: f                   \ start a new word definition
  dup 1               \ duplicate the input and put a 1 on the stack
  ?do                 \ begin counted loop from 1 to n (does nothing if 1 == n)
    i .               \ output loop index with a space appended
    i 1 .r            \ output loop index right-aligned with minimum 1 character length
  loop                \ end the loop
  .                   \ output the input we duplicated earlier
;                     \ end the word definition

Answer 48

Yabasic, 37 bytes

An anonymous function that takes input from STDIN and outputs to STDOUT.

Input""n
?1
For i=2TO n?i-1,"",i
Next

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-14 bytes thanks to @ErikF

Answer 49

Excel VBA, 29 bytes

An anonymous VBE immediate window function that takes input from cell [A1] and outputs to the console.

?1:For i=2To[A1]:?i-1 &i:Next

Answer 50

dc, 29 bytes

[d1-d0<F]dsFx+p[npz1<G]sGz1<G

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This is a good showing for dc because its printing capabilities are so limited, but line up perfectly with the challenge spec. Input is from the stack (and must be the only thing on the stack), output is to stdout. There's some duplication near the end I can't figure out how to get rid of without breaking the case n=1

Explanation

F will be the macro [d1-d0<F]
d        copy the top of the stack
 1-      decrement
   d0<F  repeat (tail recursion) until a 0 is on top

G will be the macro [npz1<G]
n        print the top (pop, but no newline)
 p       print the new top (newline, but no pop)
  z1<G   repeat (tail recursion) until only one number is on the stack

Full program (example input: 4)
[d1-d0<F]dsFx      Store F and run it: new stack is 0 1 2 3 4
+                  Drop the zero (can't just recurse to 1, or n=1 breaks)
p                  Print (no pop) the '1': stack is 1 2 3 4
[npz1<G]sG        Store G
z1<G               Run G if there's more than one number on the stack

Answer 51

dc, 34 bytes

sj1psi[lid1+dsiZAr^*li+pzlj>M]dsMx

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Completely different (and less golfy, unfortunately) approach than Sophia's dc answer. I juggle a lot of register activity here that I wish I could cut down, seems the likeliest way to golf a couple of bytes. Expects input as the sole stack entry.

sj1psi stores the input into register j, prints the initial 1, and stores 1 in register i. Macro M duplicates i, increments it, and then multiplies it by 10 to the power of however many digits the incremented value has (ZAr^*). Fetches the newly incremented version, and adds the two together. Compares register j to the number of entries on the stack, and runs until they match.

Answer 52

MATLAB 45 63 bytes

@(j)eval('for a=0:j-1;disp([num2str(a)*a/a,num2str(a+1)]);end')

Somewhat ugly *a/a in order to avoid printing leading zero for the first element

Fixed by wrapping into disp and anonymous function

Answer 53

yup, 48 bytes

0eee0ee-0ee-*:0e#[:@]0e-{]:[:]-:#0e0~--#]:@]0e-}

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Explanation

yup only knows of a few commands. To modify data, I'll be using 0 (nilad pushing that number), e (natural exponentiation), and - (subtraction).

For example, the snippet 0e is equal to 1, since \$e^0=1\$.

This program is divided into two parts: initialization and iteration.

Initialization

0eee0ee-0ee-*:0e#[:@]0e-

0eee pushes \$e^e\approx15.1543\$ and 0ee pushes \$e\approx2.7183\$. Thus, the expression 0eee0ee-0ee- pushes:

$$e^e-e-e\approx9.7177$$

This encodes 0x10 (the linefeed), since numbers are rounded to the nearest integer before being outputted. This is our numeric separator.

The next part is to initialize the stack with data. *: pushes the input twice, and 0e# outputs a 1. [:@] outputs a linefeed without popping it from the stack, and 0e- decrements the input. This handles the edge case of outputting 1 instead of 01. Rather than handle this with a conditional, hardcoding the first entry is shorter.

Iteration

{]:[:]-:#0e0~--#]:@]0e-}

{...} loops while the TOS is defined and positive. In this case, it will stop when it hits zero. Each iteration starts off with a stack like this:

[9.7177, input, iterator]

iterator starts at input - 1. ]:[:]- calculates input - iterator and gives us our true iterator value. Then, :#0e0~--# first outputs the true iterator then the same number plus 1, where 0e0~-- encodes \$-(-n-e^0)=-(-n-1)=n+1\$. Then, ]:@] restores the stack to its initial shape and 0e- subtracts 1 from the iterator, continuing our loop. This repeats until iterator reaches 0, at which point the loop stops and the program terminates.

Answer 54

D, 93 bytes

import std.math;_[]f(_)(_ n){_[]k;foreach(i;0..n)k~=++i+10^^(cast(_)i--.log10+1)*i;return k;}

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I couldn't find a nice (short) solution that evades an import, and this was the shortest of the bunch.

Answer 55

QBasic, 60 53 bytes

INPUT n
?1
FOR i=2TO n
?STR$(i-1)MID$(STR$(i),2)
NEXT

A math-based solution, because converting numbers to strings in QBasic is a lot more complicated than it ought to be. Nope, strings are still shorter, you just have to special-case the first item. The problem with STR$(i) is that it adds a space to the start of positive numbers, so we take all but the first character by using MID$. Other than that, it's pretty straightforward.

Answer 56

DC, 33 bytes

si[lidZAr^spd1-dsilp*+li0<d]dsdxf

Explanation

si[lidZAr^spd1-dsilp*+li0<d]dsdxf  Whole program.
si                                 Save the value on the stack to the i register.
  [                        ]dsdx   Create a macro, duplicate it, store it in the d register, and execute it.
   li                              Put the value from the i register on the stack.
     dZ                            Duplicate the top value, and change it to the number of digits it has.
       Ar                          Push 10 on the stack, and reverse the top two values.
         ^sp                       Do 10^x, and store it in the p register.
            d1-                    Duplicate the top value, and subtract 1 from it.
               dsi                 Duplicate the top value, and store it in the i register.
                  lp*+             Put the value from the p register on the stack, and multiply it by the the value we just put in the i register. Then, add the top two values together.
                      li0<d        Put the value from the i register and then 0 on the stack. If 0 < li, then we run the d macro again.
                                f  Print the stack.

Input is the number of numbers to be generated, on the stack.

Output is printed.

Answer 57

Kotlin, 48 bytes

{"1, "+(2..it).map{"$it${it+1}"}.joinToString()}

Yes I know, I could save 1 more byte by removing the blank behind the first comma, but it looks nicer with it :-)

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Answer 58

Shakespeare Programming Language, 310 bytes

,.Ajax,.Ford,.Page,.Act I:.Scene I:.[Enter Ajax and Ford]Ford:Listen tothy.Ajax:You cat.Open heart.[Exit Ajax][Enter Page]Scene V:.Ford:Am I worse Ajax?If notlet usScene X.You be twice the sum ofa cat a big big cat.Speak thy.Page:Open heart.You be the sum ofyou a cat.Open heart.Let usScene V.Scene X:.[Exeunt]

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Answer 59

MathGolf, 8 6 5 4 bytes

{└§p

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Outputs a newline separated series.

Explanation:

{└§p
{       Start a loop over the range(0, input)
 └      Push the top of stack (implicitly the index of the loop) + 1
  §     Concatenate the two (this removes leading 0s)
   p    And print the value

Answer 60

Runic Enchantments, 33 bytes

/01iR1-:0)?;{1+:
\$1<\}$q}:+1{$ '

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Uncompressing the entry sequence and moving the reflectors to the other side:

>1$01iR1-:0)?;{1+:\
      \}$q}:+1{$ '/

Entry sequence (>1$01i) is fairly straight forward. Push and print 1, push 0, push 1, read input and push it to the top of the stack.

At R we enter the program's main loop (unrolled with directional control characters removed):

1-:0)?;{1+:' ${1+:}q$}

At this point the stack is [0,1,i] where i is the input value.

The loop subtracts 1 from the input value (1-), compares it to greater than 0 (if true, skip terminator, else terminate; :0)?;).

Then a series of stack manipulations ({1+:{1+:}) and increments to result in [2,(i-1),1,1,2] as well as printing a space (' $). q then concats the top two items on the stack, which is then printed (giving 12 in the output stream).

Finally the stack is rotated once more, leaving [1,2,(i-1)] as the input to the next loop iteration.

Bonus challenge: using two IPs? 40 bytes

>1$0iR1-:0)?;1{+:' \
> F1iU }$~?=am$?=9m/

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As there's no way to clone the input to a second instruction pointer (I have thought about stack cloning, so this may be possible in the future, but the spec for it would be difficult to implement), we have to read it from the input stream twice.

Flow results in the second pointer being a step behind the first (avoiding merging) and the Fizzle lets us distinguish the two IPs, letting one print a space and the other discards it. I can't figure out a shorter way of performing this check.

However if it allowable to print two spaces as a separator it can be reduced to this (30 bytes):

>1$0iR1-:0)?;1\
> F1iU}$:+{$ '/

input: 4 4
output: 1  12  23  34

But this is an admittedly dubious answer due to how it has to take input, but 3 bytes shorter than the single IP answer, which is interesting.

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Update: Stack transfer

Getting the two pointers to enter the T command in the right execution order is a huge pain. The remaining two spaces in this program can't be removed, as it messes with the timing, but it avoids having to supply the input value twice. Prints 2 spaces between each entry in the sequence (35 bytes).

>1$0y TR1-:0)?;1\
 >1i:1/U}$:+{$ '/

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Answer 61

MBASIC, 102 94 bytes

1 INPUT N:FOR I=0 TO N-1:IF I>0 THEN PRINT MID$(STR$(I),2);
2 PRINT MID$(STR$(I+1),2);" ";:NEXT

Output:

? 3
1 12 23

? 10
1 12 23 34 45 56 67 78 89 910

? 16
1 12 23 34 45 56 67 78 89 910 1011 1112 1213 1314 1415 1516

Could have been much cleaner, but the STR$(n) number-to-string conversion function returns with a leading space that had to be dealt with.

Turns out the variable for NEXT and the trailing PRINT are not needed, saving 8 bytes.

Answer 62

><>, 15 13 bytes

ln1-:?!;aoln:

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We take in the input as a command line argument.

Thanks to @jo king for 2 byte loss (noticing the input loop value could be used as stack length.

Explanation (simple):

ln1-:?!;aoln:
ln              : Add the length of the stack to the stack and print.
   1-:?!;       : Take 1 off the input loop and check if zero, if 0, end.
         ao     : Print new line
           ln:  : Add the length of the stack to the stack, print and duplicate the input loop.

Answer 63

Jelly, 8 6 bytes

ṭ’DFḌ)

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ṭ’DFḌ)    Monadic link with argument n
     )    For each int in the range 1..n
 ’        n-1
ṭ         [n-1, n]
  D       Digits of n-1 and n
   F      Flatten into one list
    Ḍ     Convert back to an integer.

Answer 64

Tcl, 59 bytes

proc C n {puts [incr i]
time {puts $i[incr i]} [expr $n-1]}

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PS: I will golf it more later!

Answer 65

Haskell, 57 55 bytes

f n=read<$>zipWith((.show).(++).show)[0..][1..n]::[Int]

A function which takes a number and returns a list.

EDIT: -2 Bytes thanks to Cat Wizard

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Answer 66

RAD, 28 27 bytes

(⊢(⊢+⊣×10*1+∘⌊10⍟⊢)1+⊢)¨0,⍳

Repository

Must be called like: ((⊢(⊢+⊣×10*1+∘⌊10⍟⊢)1+⊢)¨0,⍳) <value>, as I broke assignment of functions when I implemented shy results.

Note: this should also work in Dyalog APL, and due to that: -1 byte thanks to @Adám!

The ⍳ can either be the APL symbol or the greek one if run in RAD, but must be the APL version ifran in Dyalog APL

Answer 67

CJam, 11 bytes

{{_)s+si}%}

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Answer 68

Python 2, 43 bytes

lambda i:[1]+[`x`+`x+1`for x in range(1,i)]

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First element is an integer, the rest are strings.

Answer 69

Perl 5, 29 bytes

say"@{[1,map$_-1 .$_,2..<>]}"

Answer 70

Julia 0.6, 30 bytes

n->["1";["$i$(i+1)"for i=1:n]]

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Alternative version with numeric output:

34 bytes

n->[10^ndigits(i)*~-i+i for i=1:n]

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Answer 71

C# (Visual C# Compiler), 54 bytes

Uses space as seperator character.

n=>{var r="";for(;n-->1;)r=" "+n+(n+1)+r;return"1"+r;}

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Ungolfed full code:

class P
{
    static void Main()
    {
        System.Func<int, string> f =
            n =>
            {
                var r = "";        //Declare variable for the result
                for (; n--> 1;)    //Loop until n is 1

                    r =            //Set the result to:
                        " " +      //Seperator +
                        n +        //Value of n as string
                        (n + 1) +  //Value of n + 1 as string
                        r;         //Previous content of the result

                return "1" + r;    //Return 1 (first item) + the result.
            }
            ;

        System.Console.WriteLine(f(1));
        System.Console.WriteLine(f(2));
        System.Console.WriteLine(f(3));
        System.Console.WriteLine(f(10));
        System.Console.WriteLine(f(100));
    }
}

Answer 72

C# (Visual C# Compiler), 46 bytes

If you are allowed to have some using static directives in the top whose bytes do not count:

n=>1+Concat(Range(1,n-1).Select(x=>","+x+++x))

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Will explode if input n is 0.

It is interesting that, while the C# rules say ","+x+++x means ( "," + (x++) ) + x, if it had meant ( "," + x ) + (++x) instead, the program would still work!

Answer 73

Clojure, 54 bytes

#(reduce(fn[a n](conj a(str n(inc n))))[1](range 1 %))

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Just a reduction over a range between 1 (inclusive) and n (exclusive).

(defn concat-n-n+1 [n]
  (reduce (fn [acc m]
            (conj acc (str m (inc m))))
          [1]
          (range 1 n)))

The first element of the list is a number, the rest are strings. This seems to be allowed by the spec.

Answer 74

K (oK), 12 bytes

Solution:

,/'$1,2':1+!

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Examples:

,/'$1,2':1+!1
,,"1"
,/'$1,2':1+!2
(,"1"
 "12")
,/'$1,2':1+!3
(,"1"
 "12"
 "23")
,/'$1,2':1+!10 
(,"1"
 "12"
 "23"
 "34"
 "45"
 "56"
 "67"
 "78"
 "89"
 "910")

Explanation:

,/'$1,2':1+! / the solution
           ! / range 0..n
         1+  / add 1
      2':    / sliding window of 2
    1,       / prepend 1
   $         / convert to strings
,/'          / flatten each

Answer 75

C Sharp 51 bytes

n=>new int[n].Select((_,o)=>int.Parse($"{o}{o+1}"))

Answer 76

Attache, 15 bytes

{N!_'-~_}=>Iota

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Explanation

{N!_'-~_}=>Iota
{       }=>        map the inside function:
           Iota    (over each number k from 0 to n-1)
    '              array concatenate:
   _               k
     -~_           k+1
 N!                convert [k, k + 1] to an integer (concatenates and returns a number)

Answer 77

Rutger, 108 bytes

x=$Input;
f=For[$x];
f=f[@i];
f=f[{p=Subtract[$i];r=Concat[Str[p[1]]];Print[Integer[r[Str[$i]]]];}];
Do[$f];

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Ungolfed

input = $Input;
for = For[$input];
for = for[@index];

for = for[{
    dec = Subtract[$index];
    ret = Concat[Str[dec[1]]];
    Print[Integer[ret[Str[$index]]]];
}];

Do[$for];

The basic concept of Rutger is the lack of multi-adic commands: every command takes a single argument, and if the command needs multiple argument, each new call returns a curried function.

First, we take evaluated input and store it in \$x\$. We then create a For object, a tri-adic command. The first two calls create a variable \$f\$, that will loop \$x\$ times, using the iteration variable \$i\$, starting at \$i := 1\$. The third call indicates the code to be run:

{p=Subtract[$i];r=Concat[Str[p[1]]];Print[Integer[r[Str[$i]]]];}

This creates a block of code containing 3 statements. Our first two are variable assignments. First, we create a curried function Subtract[$i], which prepares to subtract a value from \$i\$. This curried function is saved in the \$p\$ variable. Next, we create a second curried variable, \$r\$, which starts by calling p[1], subtracting \$1\$ from \$i\$. We then convert \$i-1\$ to a string and pass it as the first argument to the Concat function. Finally, we convert \$i\$ to a string, concatenate it to \$str(i - 1)\$ and convert it to an integer to reomve the leading \$0\$ when \$i = 1\$. This is then printed.

The last line, Do[$f];, runs the for loop.

Answer 78

C, 39 bytes

x(n){n&&x(n-1);printf("%d%d, ",n,n+1);}

eliminated the ternary conditional and shaved off 3 bytes.

Answer 79

ARBLE, 23 20 bytes

Saved a handful of bytes by using better named variables.

range(0,n)//(x..y|0)

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Answer 80

RAD, 72 bytes

0,(⊢+10×⊣)⍁¨∊¨(⌽∘⌊(⊢|⍨10*(⍳(1+∘⌊10⍟⊢)))÷10*1-⍨(⍳(1+∘⌊10⍟⊢)))¨¨(⊢,1+⊢)¨⍳⎕

Link to repository

I wanted to try to see if I could do it in a "non-mathematical" way by splitting a number up into its digits.

Answer 81

Ruby, 34 33 bytes

->n{$><<i||=p(1);p(i+=1)<n&&redo}

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Pseudocode:

loop
  if i is undefined
    set i to 1
    print 1
    print newline
  end if
  print i
  increment i by 1
  print i
  print newline
  if i >= n
    break and return
  end if
end loop

Answer 82

Python 2.7, 52 bytes

I tried my best. Not yet familiar with this golfing thing.

def a(n):
    for e in range(n):
        print int(`e`+`e+1`)

The int() removes the leading zero in the first output, as specified in the rules.

Answer 83

Twig, 72 bytes

Twig is very verbose, causing some issues when trying to reduce the length.

{%macro f(a)%}{%for i in 1..a%}{{o~i}}
{%set o=i%}{%endfor%}{%endmacro%}

This requires that "strict variables" is disabled (default).

---

How to use?

Simply import the macro and call it:

{% import "fn.twig" as fn %}
{{ fn.f(<number>) }}

You can test it on https://twigfiddle.com/lah1a5

Answer 84

Haskell, 35 bytes

f n="1":map(show.(-1+)<>show)[2..n]

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Explanation / Ungolfed

The operator (<>) is the addition of Semigroups, in case of the Semigroup a -> b (where b needs to be a Semigroup) it is defined as:

(f <> g) = \x-> f x <> g x

And in case of the Semigroup String it is the same as concatenation, so the code becomes:

f n = "1" : map (\x-> show (x-1) ++ show x) [2..n]

---

_{1 (imports (<>) since it's not part of the Prelude in GHC 8.2.2)}

Answer 85

TI-Basic, 22 bytes

:seq(A,A,1,Ans
:Ans+(Ans-1)10^(1+int(log(Ans

Answer 86

Bash, 56 bytes

echo 1;for i in $(seq $(($1-1)));do echo $i$((i+1));done

A pretty naive approach.

Answer 87

Lua, 62 bytes

loadstring't={1}for i=2,(...)do t[#t+1]=(i-1)..i end return t'

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---

Explanation

This is an anonymous function.

t={1} -- initializes the return table, with the number 1 already in it
for i = 2, (...) do -- loop from 2 to the number of the input
                  -- (this is actual code, ... gets the arguments of the program/function
  t[#t+1] = (i-1)..i -- append to the table i-1 concatenated with i
end
return t -- returns the table

Answer 88

Swift 4, 43 bytes

print(1);(1..<n).map{print("\($0)\($0+1)")}

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n is the input of the program

Answer 89

Noether, 15 bytes

I(iWi1+W+WP?!i)

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Explanation

I(             ) - Loop until the top of the stack equals the input
  iW             - Push i and convert it to a string
    i1+W         - Add one to i and convert to string
        +        - Concatenate two strings
         WP      - Convert string to a number and print it
           ?     - Print a newline
            !i   - Increment i

Answer 90

VBA (Excel), 31 bytes

using immediate window and Cell [A1] as input

for x=1to[a1]:?int(x-1 &x):next

Answer 91

Pepe, 68 bytes

RrEEEEEREeErREEREEEEEREErEEEEEeEErEEEereEEreeErEEEEEeeEReerREEREeRee

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Previous code has errors when giving input 0, so this is fixed!

Explanation:

# Preparation

  RrEEEEE # Stack r: [0, 1]

  REeE # Stack R: [input]

# Loop

  rREE # create label input (r flag: skip until REe)

    REEEEE # increment input (loop preparation)
    REE # create label input

      rEEEEEeEE # join all
      rEEEe # move r pointer to last
      reEE # output as int
      reeE # output newline "\n"
      rEEEEEeeE # increment all

    Ree # repeat if input != last content of stack
    rREE # create label input + 1 (r flag: skip until REe)
    REe # stop skipping commands

  Ree # if not 0, go to loop

Answer 92

Python 2, 42 bytes

I know there are plenty of shorter answers already.

a=0
exec"print`a`[:a]+`a+1`;a+=1;"*input()

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Answer 93

Kotlin, 43 bytes

Lambda takes an Int and returns a List<Int> representing the sequence.

{(1..it).mapIndexed{i,v->(""+i+v).toInt()}}

Explanation

{                                                   // lambda (Int) -> List<Int>
    (1..it)                                         // range from 1 to input
           .mapIndexed { i, v ->                    // map over "i"ndex and "v"alue
                                 (""+i+v)           // coerce to string and concat
                                         .toInt()}  // convert to int

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Answer 94

Microscript II, 16 bytes

1vsN-s{lPps1+v}*

Output is newline-separated.

Answer 95

Gol><>, 11 bytes

IFLL?nLPN|;

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Explanation:

IFLL?nLPN|;

I           //Take a number as input
 F          //Loop as many times as the input specified
  LL        //  Push the loop counter twice to the stack (same as L:)
    ?n      //  Check wether the count is zero, if not print the counter as the first digit
      LP    //  Push the loop counter and add 1
        N   //  Output the next digits of the number with nl
         |; //Exit code

Answer 96

jq, 30 characters

range(.)|"\(.)\(.+1)"|tonumber

Sample run:

bash-4.4$ jq 'range(.)|"\(.)\(.+1)"|tonumber' <<< 5
1
12
23
34
45

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Answer 97

Dart, 46 45 bytes

f(n)=>List.generate(n,(e)=>e<1?1:'$e${e+1}');

-1 byte by replacing e==0 by e<1

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